[tex]y=\dfrac{sin(x^2)}{x^3}[/tex]
Apply the quotient rule:
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{x^3\frac{\mathrm d\sin(x^2)}{\mathrm dx}-\sin(x^2)\frac{\mathrm dx^3}{\mathrm dx}}{(x^3)^2}[/tex]
Chain and power rules:
[tex]\dfrac{\mathrm d\sin(x^2)}{\mathrm dx}=\cos(x^2)\dfrac{\mathrm dx^2}{\mathrm dx}=2x\cos(x^2)[/tex]
Power rule:
[tex]\dfrac{\mathrm dx^3}{\mathrm dx}=3x^2[/tex]
Putting everything together, we have
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{x^3(2x\cos(x^2))-\sin(x^2)(2x\cos(x^2))}{(x^3)^2}[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x^3\cos(x^2)-2x\sin(x^2)\cos(x^2)}{x^6}[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x^2\cos(x^2)-\sin(2x^2)}{x^5}[/tex]
When [tex]x=\sqrt{\frac\pi2}[/tex], we have
[tex]\cos\left(\left(\sqrt{\dfrac\pi2}\right)^2\right)=\cos\left(\dfrac\pi2\right)=0[/tex]
[tex]\sin\left(2\left(\sqrt{\dfrac\pi2}\right)^2\right)=\sin(\pi)=0[/tex]
so the derivative is 0.
